BST Delete Operation in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output after deleting a leaf node in BST

What is the printed state of the BST after deleting the leaf node with value 2?

DSA Python

class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None

def insert(root, key):
    if root is None:
        return Node(key)
    if key < root.val:
        root.left = insert(root.left, key)
    else:
        root.right = insert(root.right, key)
    return root

def inorder(root, res):
    if root:
        inorder(root.left, res)
        res.append(root.val)
        inorder(root.right, res)

def deleteNode(root, key):
    if root is None:
        return root
    if key < root.val:
        root.left = deleteNode(root.left, key)
    elif key > root.val:
        root.right = deleteNode(root.right, key)
    else:
        if root.left is None:
            return root.right
        elif root.right is None:
            return root.left
        temp = root.right
        while temp.left:
            temp = temp.left
        root.val = temp.val
        root.right = deleteNode(root.right, temp.val)
    return root

root = None
for v in [5, 3, 7, 2, 4]:
    root = insert(root, v)
root = deleteNode(root, 2)
result = []
inorder(root, result)
print(result)

A [3, 4, 5, 7]

B [2, 3, 4, 5, 7]

C [3, 5, 7]

D [4, 5, 7]

💡 Hint

Deleting a leaf node removes it without affecting other nodes.

Node 2 is a leaf and gets removed. The inorder traversal then shows [3, 4, 5, 7].

❓ Predict Output

intermediate

Output after deleting a node with one child in BST

What is the printed state of the BST after deleting the node with value 3 which has one child?

DSA Python

class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None

def insert(root, key):
    if root is None:
        return Node(key)
    if key < root.val:
        root.left = insert(root.left, key)
    else:
        root.right = insert(root.right, key)
    return root

def inorder(root, res):
    if root:
        inorder(root.left, res)
        res.append(root.val)
        inorder(root.right, res)

def deleteNode(root, key):
    if root is None:
        return root
    if key < root.val:
        root.left = deleteNode(root.left, key)
    elif key > root.val:
        root.right = deleteNode(root.right, key)
    else:
        if root.left is None:
            return root.right
        elif root.right is None:
            return root.left
        temp = root.right
        while temp.left:
            temp = temp.left
        root.val = temp.val
        root.right = deleteNode(root.right, temp.val)
    return root

root = None
for v in [5, 3, 7, 4]:
    root = insert(root, v)
root = deleteNode(root, 3)
result = []
inorder(root, result)
print(result)

A [3, 4, 5, 7]

B [4, 5, 7]

C [5, 7]

D [4, 7]

💡 Hint

Deleting a node with one child replaces it with its child.

Node 3 has one child 4. After deletion, 4 takes 3's place. Inorder traversal is [4, 5, 7].

❓ Predict Output

advanced

Output after deleting a node with two children in BST

What is the printed state of the BST after deleting the node with value 3 which has two children?

DSA Python

class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None

def insert(root, key):
    if root is None:
        return Node(key)
    if key < root.val:
        root.left = insert(root.left, key)
    else:
        root.right = insert(root.right, key)
    return root

def inorder(root, res):
    if root:
        inorder(root.left, res)
        res.append(root.val)
        inorder(root.right, res)

def deleteNode(root, key):
    if root is None:
        return root
    if key < root.val:
        root.left = deleteNode(root.left, key)
    elif key > root.val:
        root.right = deleteNode(root.right, key)
    else:
        if root.left is None:
            return root.right
        elif root.right is None:
            return root.left
        temp = root.right
        while temp.left:
            temp = temp.left
        root.val = temp.val
        root.right = deleteNode(root.right, temp.val)
    return root

root = None
for v in [5, 3, 7, 2, 4]:
    root = insert(root, v)
root = deleteNode(root, 3)
result = []
inorder(root, result)
print(result)

A [2, 3, 4, 5, 7]

B [4, 5, 7]

C [2, 4, 5, 7]

D [2, 5, 7]

💡 Hint

Deleting a node with two children replaces it with its inorder successor.

Node 3 has two children. It is replaced by its inorder successor 4. Inorder traversal is [2, 4, 5, 7].

🔧 Debug

advanced

Identify the error in BST delete function

What error will this code produce when deleting a node with two children?

DSA Python

def deleteNode(root, key):
    if root is None:
        return root
    if key < root.val:
        root.left = deleteNode(root.left, key)
    elif key > root.val:
        root.right = deleteNode(root.right, key)
    else:
        if root.left is None:
            return root.right
        elif root.right is None:
            return root.left
        temp = root.right
        while temp.left:
            temp = temp.left
        root.val = temp.val
        root.right = deleteNode(root.right, temp.val)
    return root

A No error; code works correctly

B Infinite loop due to wrong traversal in finding successor

C AttributeError because temp might be None

D Deletes wrong node because it finds the rightmost instead of the leftmost node in the right subtree

💡 Hint

Check how the successor node is found in the right subtree.

The code correctly finds the leftmost node in the right subtree (inorder successor). This is the correct approach.

🧠 Conceptual

expert

Number of nodes after multiple deletions in BST

Given a BST with nodes inserted in this order: [10, 5, 15, 3, 7, 12, 18], after deleting nodes 5 and 15, how many nodes remain in the BST?

A 7

B 6

C 4

D 5

💡 Hint

Each deletion removes exactly one node.

Initially 7 nodes. Deleting 5 and 15 removes 2 nodes. Remaining nodes = 7 - 2 = 5.

Practice

1. What happens when you delete a leaf node (a node with no children) from a Binary Search Tree (BST)?

easy

A. The node's children are moved up to replace it.

B. The entire tree is rebuilt from scratch.

C. The node is simply removed without affecting other nodes.

D. The root node is always changed.

Solution

1. Step 1: Understand leaf node deletion

   A leaf node has no children, so removing it does not affect other nodes.

2. Step 2: Effect on BST structure

   Since no children exist, simply remove the node and update its parent's pointer to null.

3. Final Answer:

   The node is simply removed without affecting other nodes. -> Option C

4. Quick Check:

   Leaf node deletion = simple removal [OK]

Hint: Leaf nodes delete by direct removal only [OK]

Common Mistakes:

• Thinking children need rearrangement
• Assuming root always changes
• Believing tree rebuild is needed

2. Which of the following is the correct Python function signature to delete a node with a given key from a BST?

easy

A. def deleteNode(key):

B. def deleteNode(root, key):

C. def deleteNode(root):

D. def deleteNode(key, root):

Solution

1. Step 1: Identify required parameters

   To delete a node, we need the BST root and the key to delete.

2. Step 2: Check function signature

   Only def deleteNode(root, key): has both parameters in correct order.

3. Final Answer:

   def deleteNode(root, key): -> Option B

4. Quick Check:

   Function needs root and key [OK]

Hint: Delete needs root and key parameters [OK]

Common Mistakes:

• Swapping parameter order
• Missing one parameter
• Using only key or root alone

3. Given the BST below, what is the inorder traversal after deleting node 10?

       15
      /  \
    10    20
   /  \     \
  8   12    25

medium

A. [8, 12, 15, 20]

B. [8, 10, 12, 15, 20, 25]

C. [8, 15, 20, 25]

D. [8, 12, 15, 20, 25]

Solution

1. Step 1: Delete node 10 with two children

   Node 10 has two children (8 and 12). Replace it with its inorder successor (12).

2. Step 2: Adjust tree and perform inorder traversal

   After replacement, node 12 replaces 10, and 12's original position is removed. Inorder traversal is [8, 12, 15, 20, 25].

3. Final Answer:

   [8, 12, 15, 20, 25] -> Option D

4. Quick Check:

   Inorder after delete = [8, 12, 15, 20, 25] [OK]

Hint: Replace deleted node with inorder successor [OK]

Common Mistakes:

• Not replacing node with successor
• Leaving deleted node in traversal
• Incorrect traversal order

4. The following code attempts to delete a node from a BST but causes an error. What is the main issue?

def deleteNode(root, key):
    if not root:
        return root
    if key < root.val:
        deleteNode(root.left, key)
    elif key > root.val:
        deleteNode(root.right, key)
    else:
        if not root.left:
            return root.right
        elif not root.right:
            return root.left
        temp = findMin(root.right)
        root.val = temp.val
        deleteNode(root.right, temp.val)
    return root

medium

A. Recursive calls do not update root's children pointers.

B. The base case is missing.

C. The function does not handle leaf nodes.

D. The function uses an undefined findMin function.

Solution

1. Step 1: Check recursive calls for updating children

   Recursive calls like deleteNode(root.left, key) do not assign the result back to root.left, so changes are lost.

2. Step 2: Correct by assigning returned subtree

   We must write root.left = deleteNode(root.left, key) and root.right = deleteNode(root.right, key) to update pointers properly.

3. Final Answer:

   Recursive calls do not update root's children pointers. -> Option A

4. Quick Check:

   Update children with recursive call results [OK]

Hint: Assign recursive call results to children [OK]

Common Mistakes:

• Ignoring returned subtree after recursion
• Assuming in-place changes without assignment
• Missing base case (not the main issue here)

5. You want to delete a node with two children in a BST. Which approach correctly maintains BST properties after deletion?

hard

A. Replace the node with its inorder predecessor or successor, then delete that node.

B. Replace the node with its left child only.

C. Replace the node with its right child only.

D. Remove the node and leave a gap in the tree.

Solution

1. Step 1: Understand deletion of node with two children

   Direct removal breaks BST rules; we must replace the node with a suitable value.

2. Step 2: Use inorder predecessor or successor

   Replacing with inorder predecessor (max in left subtree) or successor (min in right subtree) keeps BST order intact.

3. Step 3: Delete the replaced node recursively

   After replacement, delete the predecessor/successor node which will have at most one child.

4. Final Answer:

   Replace the node with its inorder predecessor or successor, then delete that node. -> Option A

5. Quick Check:

   Use inorder predecessor/successor for two-child deletion [OK]

Hint: Use inorder predecessor or successor for two-child node [OK]

Common Mistakes:

• Replacing with only left or right child
• Removing node without replacement
• Breaking BST ordering rules